sql - 查找丢失的序列号

Question

我想创建一个表来查找丢失的序列号。0到70000之间的序列号达到70000后变为0。在特定的时间段内我需要找到那些丢失的记录。

Answer 1

此解决方案基于生成从 1 到您设置的某个限制的所有自然数的语句：

SELECT ROWNUM N FROM dual CONNECT BY LEVEL <= 7000

该解决方案的第二部分是 OracleMINUS运算符（通常称为EXCEPT），旨在减去集合。

换句话说，最终查询是：

SELECT ROWNUM id FROM dual CONNECT BY LEVEL <= 7000
MINUS
SELECT id FROM mytable

20 个数字的 SQLFiddle 演示。

Answer 2

您可以使用Lead 和 lag 函数来检测序列中的间隙。
该解决方案不会限制您使用特定的上限数字，例如 70000。

检测：

SELECT *
  FROM (SELECT lag(c.id) over(ORDER BY id) last_id,
               c.id curr_id,
               lead(c.id) over(ORDER BY id) next_id
          FROM mytable c
         order by id)
 WHERE nvl(last_id, curr_id) + 1 <> curr_id
   AND last_id IS NOT NULL

Sqlfiddle 演示。

遍历：

begin
  FOR x IN (SELECT *
              FROM (SELECT lag(c.id) over(ORDER BY id) last_id,
                           c.id curr_id,
                           lead(c.id) over(ORDER BY id) next_id
                      FROM mytable c order by id)
             WHERE nvl(last_id, curr_id) + 1 <> curr_id AND 
             last_id IS NOT NULL
            ) LOOP
    dbms_output.put_line('last_id :' || x.last_id);
    dbms_output.put_line('curr_id :' || x.curr_id);
    dbms_output.put_line('next_id :' || x.next_id);
    dbms_output.put('gaps found: ');
    for j in x.last_id + 1 .. nvl(x.next_id,x.curr_id) - 1   loop
      if  j != x.curr_id then
      dbms_output.put(j || ', ');
      end if;      
    end loop;
    dbms_output.put_line('');
    dbms_output.put_line('*****');
  end loop;
end;

Answer 3

不久前我从 Tom Kyte 那里偷了这个：

select  id, one_before, Diff, dense_rank() over (order by Diff desc) rank from (
select  id, one_before,
            case when (id - one_before) > 1 then (id - one_before)
           else 1
           end Diff
 from (
       select id, lag(id) over(order by id) one_before
      from table_name order by id) )

原始讨论位于http://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:8146178058075。